Question: Divide the following complex numbers. $ \dfrac{-45-5i}{4+5i}$
Answer: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${4-5i}$ $ \dfrac{-45-5i}{4+5i} = \dfrac{-45-5i}{4+5i} \cdot \dfrac{{4-5i}}{{4-5i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(-45-5i) \cdot (4-5i)} {(4+5i) \cdot (4-5i)} = \dfrac{(-45-5i) \cdot (4-5i)} {4^2 - (5i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(-45-5i) \cdot (4-5i)} {(4)^2 - (5i)^2} = $ $ \dfrac{(-45-5i) \cdot (4-5i)} {16 + 25} = $ $ \dfrac{(-45-5i) \cdot (4-5i)} {41} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-45-5i}) \cdot ({4-5i})} {41} = $ $ \dfrac{{-45} \cdot {4} + {-5} \cdot {4 i} + {-45} \cdot {-5 i} + {-5} \cdot {-5 i^2}} {41} $ Evaluate each product of two numbers. $ \dfrac{-180 - 20i + 225i + 25 i^2} {41} $ Finally, simplify the fraction. $ \dfrac{-180 - 20i + 225i - 25} {41} = \dfrac{-205 + 205i} {41} = -5+5i $